SOLUTION: The length of a conference room is one and a half times its width.A carpet is placed in the center of the room.The length of the carpet is twice it width. This leaves a 3m wide bo

Algebra ->  Equations -> SOLUTION: The length of a conference room is one and a half times its width.A carpet is placed in the center of the room.The length of the carpet is twice it width. This leaves a 3m wide bo      Log On


   

Question 1129698: The length of a conference room is one and a half times its width.A carpet is placed in the center of the room.The length of the carpet is twice it width. This leaves a 3m wide border around the edges of the carpet.Find the area of the carpet.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let +w+ = the width of the room
+%283%2F2%29%2Aw+ = the length of the room
+w+-+2%2A3+=+w+-+6+ is the width of the carpet
+%283%2F2%29%2Aw+-+2%2A3+=+%283%2F2%29%2Aw+-+6+ is the length of the carpet
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I can now say:
+%283%2F2%29%2Aw+-+6+=+2%2A%28+w+-+6+%29+
+3w+-+12+=+4w+-+24+
+w+=+12+
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+w+-+6+=+12+-+6+
+w+-+6+=+6+
and
+%283%2F2%29%2A12+-+6+=+18+-+6+
+%283%2F2%29%2A12+-+6+=+12+
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The area of the rug is:
+6%2A12+=+72+
72 m2
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check: the dimensions of the room are:
+w+=+12+ and
+%283%2F2%29%2A12+=+18+
+12+-+2%2A3+=+6+
+18+-+2%2A3+=+12+
OK